Let's solve the problem step by step: --- ### **Step 1: Write the general Fourier series formula and identify the coefficients** For a function \( f(x) \) with period \( 2\pi \): \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left[a_n \cos(nx) + b_n \sin(nx)\right] \] where, \[ a_0 = \frac{1}{2\pi} \int_{0}^{2\pi} f(x) \, dx \] \[ a_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \cos(nx) \, dx \] \[ b_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \sin(nx) \, dx \] **Explanation:** We start by writing the Fourier series formula and the formulae for the coefficients, which we need to compute for \( f(x) = x^2 \). --- ### **Step 2: Compute the first three non-zero terms** #### **Calculate \( a_0 \):** \[ a_0 = \frac{1}{2\pi} \int_{0}^{2\pi} x^2 \, dx = \frac{1}{2\pi} \left[ \frac{x^3}{3} \right]_0^{2\pi} = \frac{1}{2\pi} \cdot \frac{(2\pi)^3}{3} = \frac{1}{2\pi} \cdot \frac{8\pi^3}{3} = \frac{4\pi^2}{3} \] #### **Calculate \( a_n \):** \[ a_n = \frac{1}{\pi} \int_{0}^{2\pi} x^2 \cos(nx) \, dx \] This integral can be computed by parts or by using a standard result (see explanation at the end). The result is: \[ a_n = \frac{4}{n^2}(-1)^n \] #### **Calculate \( b_n \):** \[ b_n = \frac{1}{\pi} \int_{0}^{2\pi} x^2 \sin(nx) \, dx \] Since \( x^2 \sin(nx) \) is an odd function over \( (0, 2\pi) \), and its integral over a full period is zero: \[ b_n = 0 \] So, the Fourier series is: \[ x^2 = \frac{4\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4}{n^2}(-1)^n \cos(nx) \] The first three non-zero terms: - Constant term: \( \frac{4\pi^2}{3} \) - First cosine term: \( -4 \cos(x) \) (since \( n=1, (-1)^1 = -1 \)) - Second cosine term: \( +1 \cos(2x) \) (since \( n=2, (-1)^2 = 1 \), \( \frac{4}{4} = 1 \)) **Explanation:** We calculated the required coefficients, and observed the pattern of the series. The sine terms vanish. The first three non-zero terms are the constant, the \( \cos(x) \), and the \( \cos(2x) \) terms. --- ### **Final Answer:** \[ \boxed{ x^2 = \frac{4\pi^2}{3} - 4\cos(x) + \cos(2x) + \cdots } \] These are the first three non-zero terms of the Fourier series for \( f(x) = x^2 \) on \( (0, 2\pi) \).